metal structures design project iii steel hall examples of

Metal Structures

Design Project III

Steel hall – examples of calculation (part I)

Ist example of calculations – continous suspended cold-formed purlin

Photo: kingspan.com

Photo: Author

Dead-weigth of purlin is at now probably smaller (cold-formed is lighter than hot rolled).

But participation of dead-weight in total load (max part = snow) is small, and difference

between hot rolled and cold-formed is completely neglegible in comparison to total load.

For calculation could be taken completely the same value of load as for purlins in Ist desig

project.

yPhoto: Author

z

qz = 12,217 kN/m

qy = 0,960 kN/m

qy = qy[qH(s)] + qy[qV(s)] + qy[qH(g)] + qy[qV(g)] + qy(w) =

= 0,000 + 0,872 + 0,000 + 0,088 + 0,000 = 0,960 kN / m

qz = qz[qH(s)] + qz[qV(s)] + qz[qH(g)] + qz[qV(g)] + qz(w) =

= 0,000 + 9,962 + 0,000 + 1,000 + 1,255 = 12,217 kN / m

MEd, y = qz L2 / 8 = 54,977 kNm

MEd, z = qy L2 / 8 = 4,320 kNm

VEd, y = qy L / 2 = 2,880 kN

VEd, z = qz L / 2 = 36,651 kN

→ Des #1 examp / 10

Photo: Author

Continous multispan purlin could be calculated with using of table for calculations.

Results for 2- , 3- , 4- and 5-span bealm are presented in table. For 6- and more span we can use

information for 5-span beam.

First span

First span

Second span

Second span

Last span

Last span

Penultimate span

Penultimate span

Central span

Central span many times

Photo: Tablice do projektowania

konstrukcji metalowych, W.

Bogucki, M. Żyburtowicz, Arkady

1996 Warszawa

Photo: Tablice do projektowania konstrukcji metalowych, W. Bogucki,

M. Żyburtowicz, Arkady 1996 Warszawa

Scheme 1: for dead-weight g

Schemes 2-5: for live load q

g(1) + p(2) – max moments M1 , M3 amd reaction A

g(1) + p(3) – max moments M2

g(1) + p(4) – max moments MB and reaction B

g(1) + p(5) – max moments MC and Reaction C

Because of division between dead-wight and live loads, data from #t / 3 must be recalculated

to effects of g and q.

qH

Photo: Author

qV

s → qH, qV ; l

g → qH, qV ; l1

w → qy, qz ; l1

d1

d

qH(s) = 0,000 kN / m

qV(s) = s d = 10,000 kN / m

qH(g) = 0,000 kN / m

qV(g) = g d1 = 1,004 kN / m

qy(w) = 0,000 kN / m

qz(w) = w d1 = 1,255 kN / m

d1

d

l1 = 2 d1 / 2 = d1

l = 2 d / 2 = d


qH

Photo: Author

qV

qH → qy , qz

qV → qy , qz

qy(qH) = qH cos a

qz(qH) = - qH sin a

qy(qV) = qV sin a

qz(qV) = qV cos a

qy[qH(s)] = qH(s) cos a = 0,000 kN / m

qz[qH(s)] = - qH(s) sin a = 0,000 kN / m

qy[qV(s)] = qV(s) sin a = 0,872 kN / m

qz[qV(s)] = qV(s) cos a = 9,962 kN / m

qy[qH(g)] = qH(g) cos a = 0,000 kN / m

qz[qH(g)] = - qH(g) sin a = 0,000 kN / m

qy[qV(g)] = qV(g) sin a = 0,088 kN / m

qz[qV(g)] = qV(g) cos a = 1,000 kN / m


Ultimately, information from #t / 3

qz = 12,217 kN/m qy = 0,960 kN/m

and information from #t / 8

qy[qV(g)] = qV(g) sin a = 0,088 kN / m

qz[qV(g)] = qV(g) cos a = 1,000 kN / m

can be presented as

qz = 12,217 kN/m = qz(g) + qz(q) = gz + qz = 1,000 kN / m + 11,217 kN / m

qy = 0,960 kN/m = qy(g) + qy(q) = gy + qy = 0,088 kN / m + 0,872 kN / m

Photo: Author

Purlin is supported:

• In direction z by main roof girder;

• In direction y by main roof girder and additional suspension for purlins in halg between

girders.

This types of supports make two warious static schemes in both directions.

MEd, y (qz)

MEd, z (qy)

Photo: Author

MEd, y, max in span MEd, y, max in span MEd, y, max in span

MEd, y, max over support MEd, y, max over support

MEd, z, max over support MEd, z, max over support MEd, z, max over support

MEd, z, max over support MEd, z, max over support

Such type of static scheme makes, that max values MEd, y, max – over supports on in spans –

correspond max values MEd, z, max – only over supports.

Photo: Author

Information in table about shear forces are presented for left (L) and right (P – prawy) cross-

section nest to support. Reaction R = |QL| + |QP|

QLQP

R

Photo: Author

* * *

Reactions in additional supports in direction y are loads applied to suspension bars.

Coefficents in direction z:

Point Value g(1) q(2) q(3) q(4) q(5)

A RA 0,395 0,447

1 MEd, y, 1 0,0781 0,1000

B MEd, y, B 0,1050 0,1190

RB 0,556 1,218

2 MEd, y, 2 0,0331 0,0787

C MEd, y, C 0,0789 0,1110

RC 0,979 1,167

3 MEd, y, 3 0,0462 0,0855

A B C

1 2 3

Photo: Author

Ri = agi, R g L + aqi, R q L

Mi = agi, M g L2 + aqi, M q L2

L = 6m

Values:Point Value

A RA 32,454 kN

1 MEd, y, 1 43,193 kNm

B MEd, y, B 51,834 kNm

RB 85,310 kN

2 MEd, y, 2 32,971 kNm

C MEd, y, C 47,664 kNm

RC 84,415 kN

3 MEd, y, 3 36,189 kNm

A B C

1 2 3

Photo: Author

Max M over support

Max M in span

Max R

Cold-formed Z-purlins are often joined over support by overlap. Resistance abowe

support is abut two times bigger than in span.

Photo: ruukki.com

Photo: forminometal.com

Because of this, fo calculation important is

only bending moment in span.

Point Value

A RA 32,454 kN

1 MEd, y, 1 43,193 kNm

B MEd, y, B 51,834 kNm

RB 85,310 kN

2 MEd, y, 2 32,971 kNm

C MEd, y, C 47,664 kNm

RC 84,415 kN

3 MEd, y, 3 36,189 kNm

A B C

1 2 3

Photo: Author

A B C C C C

Direction z

Direction y

Bending moment MEd, y in span 3

corresponds bending moment MEd, z

over support C.

Point Value L = 3 m No suspension, L = 6 m (for comparison)

A RA 1,274 kN 2,548 kN

1 MEd, y, 1 0,847 kNm 3,388 kNm

B MEd, y, B 1,017 kNm 4,068 kNm

RB 3,333 kN 6,667 kN

2 MEd, y, 2 0,642 kNm 2,568 kNm

C MEd, y, C 0,934 kNm 3,736 kNm

RC 3,311 kN 6,622 kN

3 MEd, y, 3 0,708 kNm 2,832 kNm

1 2 3

Photo: Author

A B C C C C

Direction y

3 3 3L = 3m

Photo: Author

Resistance of purlin could be analised in main central

axes x h or in axes y z paralell to web and flange.

y

a

b

x

h

a

z

No cooperation between purlin and

housing (for example: after many years

post erection of structure complex of

small destructions in bolts, purlins and

housing eliminates cooperation).

Cooperation between purlin and

housing (through ~ 10 years since

erecton of structure).

Calculations in axes y-x:

Z 400-3:

Wy = 180,34 cm3

Wz = 19,02 cm3

Vertical part of web = (40 – 2 ∙ 1,5 ∙ 0,3) ∙ 0,3 = 11,73 cm2

Horizontal part of flages = 2 ∙ (8,5 – 2 ∙ 1,5 ∙ 0,3) ∙ 0,3 = 4,56 cm2

Photo: pruszynski.com.pl

sy = MEd, y / Wy + MED, z / Wz = 200,627 MPa + 47,124 MPa = 247,124 MPa

ty = Vy / Avy = 7, 309 MPa

tz = Vz / Avz = 72,728 MPa

EN 1993-1-1 (6.1):

S235:

[sy / (fy gM0)]2 + 3 [(tz

2 + tz2) / (fy gM0)

2] = 1,18 > 1,0 wrong

Deflections:

Value of deflections can be approximated by one of two ways:

• Values from „normal” static calculations;

• Values from formula:

D = 0,50 [5 g L4 / (384 E J)] + 0,75 [5 q L4 / (384 E J)]

where

D - deflection; g – dead weight; q – live load; L – length of one span

→ Proj #2 / 28

According to #t / 9:gz + qz = 1,000 kN / m + 11,217 kN / m

According to #t / 19:Jy = 3653,93 cm4

Dz = 0,50 [5 gz L4 / (384 E Jy)] + 0,75 [5 qz L4 / (384 E Jy)] = 0,020 m

Dacc (#11 / 94) = L / 200 = 0,030 m

Dz / Dacc = 0,667 < 1 OK

Photo: Author

Suspension makes, that max Dz corresponds with Dy = 0. Dy is neglegible.

Dz

Dy

Reactions in additional supports in direction y are loads applied to suspension bars (#t / 12).

According to #t / 17, this force = 3,333 kN.

Round bar F 8mm; A = 0,503 cm2

Nt,Rd = A fy / gM0 = 11,812 kN

F / Nt,Rd = 0,282 < OK

Compression in suspended bars is not taken into consideration. Such slenderness members losts

stability under even small compressive force.

Photo: A. Biegus, Przyczyny przedawaryjnego stanu technicznego płatwi

hali stalowej, Budownictwo i Architektura 12 / 2013, 173-180

Photo: dromet.pl

Purlin Purlin

Hanger after buckling

There is possible compression in part of hangers. There will be permanently deformations as

the effect of buckling.

We need rigging screws to repair hangers.

→ #8 / 34

IInd example of calculations – main column in frame

Photo: Author

Forces shoul be applied in point – as

effect of contact with purlins and wall

girts.

For big number of purlins and girts,

loads could be applied as continous.

Photo: Author

Point A:

Horizontal force = wind action on wall from wall girt + horizontal compoun from wind action

on purlin

Vertical force = dead weight of wall girt + dead weight of housing panels + dead weigth of

purlin + dead weight of roofing panels + snow + vertical compoun from wind action on purlin

Point B:

Horizontal force = wind action on wall from wall girt

Vertical force = dead weight of wall girt + dead weight of housing

Point C:

Horizontal force = horizontal compoun from wind action on purlin

Vertical force = dead weigth of purlin + dead weight of roofing panels + snow + vertical

compoun from wind action on purlin

A

B

CA

B

C

General notice

Main roof girder is calculated according to the same algorithm, as main column. The same,

resistance, stability and deformation must be taken into consideration.

Resistance is checked completely the same for girger and column.

Stability for girder is calculated for specific critical lengths (→ #10 / 57).

Deformation for girder is vertical deflection of ridge.

Photo: Author

Columns: HEA 450

H = 5,45 m

Girders: HEA 450

L = 23,50 m

MEd, top = 311,6 kNm

MEd, bottom = 0,0 kNm

NEd, top = 93,4 kN

NEd, bottom = 97,6 kN

VEd, max = 57,2 kNm

Resistance and stability of main columnn

Single-storey frame

n = 1

Photo: Author

m ≥ 1

Eurocode – N.A.

Experience – general rules

C1

T:

m ≤ 1,0

R

G

F

P

D

R

C

B

A

T:

m =?

m =1

s.i+II

s.b.i

+II

Co:

m =?

Methods:Resistance

Stability:Deformations

Critical length:

→ #13 / 92

Recommended algorithm for Your design project:

n = 1

Photo: Author

m ≥ 1

R

GDC T:

m =?

*

*

*

*

** * *

*

* * *

*

*

C1

T:

m ≤ 1,0

Co:

m =?

Few additional information (→ #t / 32)

Procedure "C"

„Normal” loads are taken into consideration only.

Photo: Author → #13 / 53

„Normal” loads in procedure C means loads from dead-weight, snow, wind, imposed,

thermal, etc. In opposite to method A or B, effects from imperfection and second order

effects are not taken into consideration. Cross-sectional forces for method C are presented

in #t / 28.

In method A or B more combinations of actions are taken into consideration in

comparison to C. But analisis of instability in these two methods is simpler, than in C.

In method C, critical length of column in plane of frame must be calculated.

According to EN 1993-1-1 N.A. 9 (C1 on #t / 30), one-storey steel frame can be calculated

without second order effects and incluences of imperfection and, additionaly, as non-sway

frame. Critical length of column in non-sway frame is not greater than height of column.

If we do not make accurate calculations of m (of course m ≤ 1,0), we can taken into

consideration m = 1,0 , the same as in method B.

Critical length can be calculated by computer (m = ?). More information is presented on

#13 / 69-70. This method is accepted in Your range of project as second method, to

verification results from tables.

Single-storey frame

"Tablice do projektowania konstrukcji metalowych", W. Bogucki, M. Żyburtowicz, Arkady, Warszawa 1984

(n = m)

→ #13 / 56

Data for calculation:

h = 5,45 m

b = 23,50 m

F = area of cross-section of column = A (HEA 450) = 98,8 cm2

J0 , J = moments of inertia for cross section of left and right column = J (HEA 450) =

= 33 740 cm4

P0, P1 – equivalent vertical loads from roof; for symmetrical frame P0 = P1

n = P1 / P0 = 1,000000

c = J b / (J0 h) = 4,311927

s = 4 J / (b2 F) = 0,000247

m = {√[0,5(1 + n)]} {√[4+1,4(c + 6s) + 0,02 (c + 6s)2]} = 3,227

Critical length of column in plane of frame is equal:

5,45 m ∙ 3,227 = 17,587 m

Critical length of column in out-of-plane of frame, because of vertical bracings between

columns, is equal:

5,45 m

The same for critical length foe lateral buckling.

Photo: steelwarehousechina.com

Photo: greenterrahomes.com

Resistance must be analised in each characteristic point separately. For frame as follow:

In top point of column and in bottom point of column.

Photo: Author



NEd, top = 93,4 kN


VEd, max = 57,2 kNm

Shear force the same should be analised separately for top and bottom poin, but for sch

small value can be analised max value.

HEA 450 [mm]

h b tf tw r

440 300 21 11,5 27

A = 178,0 cm2

Jy = 63 720 cm4

Jz = 9465 cm4

Wy, pl = 3216 cm3

Wz, pl = 965,5 cm3

AV, z ≈ 92,4 cm2

MRd, y = 755,760 kNm

NRd = 4183,0 kN

VRd = 1253,658 kN

Photo: Author

Resistance: top point


NEd, top = 93,4 kN

VEd, max = 57,2 kNm

MEd, top / MRd, y = 0,412 < 1,0 OK

NEd, top / NRd = 0,022 < 1,0 OK

VEd, max / VRd = 0,046 < 1,0 OK

VEd, max / VRd = 0,046 < 0,5 → no interaction between shear force and bending moment

min ( 0,25 Npl, Rd ; 0,5 hw tw fy / gM0) = min (1045,750 ; 464,830) = 464,830 kN

NEd, top < 464,830 kN → no interaction between axial force and bending moment

Resistance: bottom point



VEd, max = 57,2 kNm

MEd, top / MRd, y = 0,0 < 1,0 OK

NEd, top / NRd = 0,023 < 1,0 OK

VEd, max / VRd = 0,046 < 1,0 OK

No bending moment → no interaction with shear force and axial force

Stability is analised for total member. This means global calculation for maximum values

of NEd, My, Ed, Mz, Ed even if they are in three different cross-sections.

MEd = MEd, top = 311,6 kNm

NEd = NEd, bottom = 97,6 kN

NEd / ( cy NRk / gM1) + kyy (My, Ed + DMy, Ed ) / ( cLT M y, Rk / gM1) +

+ kyz ( Mz, Ed + DMz, Ed ) / (M z, Rk / gM1) ≤ 1,0

NEd / ( cz NRk / gM1) + kzy (My, Ed + DMy, Ed ) / ( cLT M y, Rk / gM1) +

+ kzz ( Mz, Ed + DMz, Ed ) / (M z, Rk / gM1) ≤ 1,0

kyy , kyz , kzy , kzz

"French" method EN 1993-1-1 App. A, table A1, A2

"German" method EN 1993-1-1 App. B, table B1, B2, B3

"French" and "German" methods:

EN 1993-1-1 (6.61), (6.62)

→ #13 / 77

Interaction between flexural buckling in plane of frame (Lcr = 17,587 m, buckling about

stron axis of HEA 450) and lateral buckling (Lcr = 5,450 m):

NEd / ( cy NRk / gM1) + kyy (My, Ed + DMy, Ed ) / ( cLT M y, Rk / gM1) +

+ kyz ( Mz, Ed + DMz, Ed ) / (M z, Rk / gM1) ≤ 1,0

Interaction between flexural buckling out of plane of frame (Lcr = 5,450 m, buckling about

weak axis of HEA 450) and lateral buckling (Lcr = 5,450 m):

NEd / ( cz NRk / gM1) + kzy (My, Ed + DMy, Ed ) / ( cLT M y, Rk / gM1) +

+ kzz ( Mz, Ed + DMz, Ed ) / (M z, Rk / gM1) ≤ 1,0

No bending moment out of plane of truss → Mz, Ed = 0,0 kNm

HEA 450 Ist class of cross section, no diference between global and effective center of

gravity, no eccentricities → ez = 0,0 m ; ey = 0,0 m

DMy, Ed = NEd ez = 0,0 kNm

DMz, Ed = NEd ey = 0,0 kNm

NEd / ( cy NRk / gM1) + kyy My, Ed / ( cLT M y, Rk / gM1) ≤ 1,0

NEd / ( cz NRk / gM1) + kzy My, Ed / ( cLT M y, Rk / gM1) ≤ 1,0

Flexural buckling, axis y Ncr, y = p2 EJy / (my l0y)2 = 4 269,837 kN

Flexural buckling, axis z Ncr, z = p2 EJz / (mz l0z)2 = 6 604,602 kN

Torsional buckling Ncr, T = [p2 EJw / (mT l0T)2 + GJt] / is2 = 5 446,494 kN

Lateral buckling Mcr = is √ (Ncr, z Ncr, T) = 1793,300 kNm

lLT = √ (Wy, pl fy / Mcr) = 0,649

FLT = 0,684

cLT = min{1/[FLT + √ (FLT2 - lLT

2)] ; 1/ lLT2 ; 1,0} = 1,0

cLT, mod = 1,0

ly = √ (A fy / Ncr, y) = 0,990

Fy = 1,128

cy = 0,599

lz = √ (A fy / Ncr, z) = 0,633

Fz = 1,306

cz = 0,408

"German" method EN 1993-1-1 tab. B.3

Moment diagram Range Cmy and Cmz and CmLT

Uniform loading Concentrated loading

-1 ≤ Y ≤ 1 max (0,6 + 0,4 Y ; 0,4)

For members with sway buckling mode, the equivalnet uniform moment factor should be taken Cmy = 0,9 or Cmz = 0,9 respectively.

Cmy, Cmz and CmLT should be obtained according to the bending moment between the relevant braced points as follows:

Moment factor: bending axis: points braced in direction:

Cmy y-y z-z

Cmz z-z y-y

CmLT y-y y-y

Y = 0

Cmy = Cmz = CmLT = max (0,6 + 0,4 Y ; 0,4) = 0,6

But „For members with sway buckling mode…” → This is sway

mode of instability for such frame.

Cmy = Cmz = 0,9

CmLT = 0,6Photo: Author

For "German" method, members are divided into susceptible to torsional deformations and

not susceptible to torsional deformations.

Cross-section Comments

No lateral buckling, no need for buckling

interaction analysis

Member not susceptible to torsional

deformations, EN 1993-1-1 tab. B.1

Member susceptible to torsional

deformations, EN 1993-1-1 tab. B.2

Short member:

1 / [FLT + √ (FLT2 - lLT

2)] ≥ 1,0

1 / [FLT + √ (FLT2 - lLT

2)] < 1,0

kq →

→ #13 / 79

Analised case → Tab B.1

EN 1993-1-1 tab. B.1

ny = NEd gM1 / (cy NRd )nz = NEd gM1 / (cz NRd )

Interaction

factors

Cross-

section

Ist, IInd class IIIrd, IVth class

kyy I, H, RHS Cmy ∙ min {1 + 0,6 ly ny ;

1 + 0,6 ny}

Cmy ∙ min {1 + (ly – 0,2) ny ;

1 + 0,8 ny}

kyz I, H, RHS 0,6 kzz kzz

kzy I, H, RHS 0,6 kyy 0,8 kyy

kzz I, H Cmz ∙ min {1 + (2 lz - 0,6) nz ;

1 + 1,4 nz} Cmz ∙ min {1 + 0,6 lz nz ;

1 + 0,6 nz}RHS Cmz ∙ min {1 + (lz - 0,2) nz ;

1 + 0,8 nz}

For I- and H-sections and RHS, under axial compression and unaxial bending My, ED, the

coefficient kzy may be = 0

"German" method,

interaction factors→ #13 / 80

ny = NEd gM1 / (cy NRd ) = 0,038

nz = NEd gM1 / (cz NRd ) = 0,056

Interaction

factors

Cross-

section

Ist, IInd class

kyy I, H, RHS Cmy ∙ min {1 + 0,6 ly ny ;

1 + 0,6 ny} = 0,919

For I- and H-sections and RHS, under axial

compression and unaxial bending My, ED, the

coefficient kzy may be = 0

NEd / ( cy NRk / gM1) + kyy My, Ed / ( cLT M y, Rk / gM1) ≤ 1,0

NEd / ( cz NRk / gM1) ≤ 1,0

97,6 / (0,599 ∙ 4183,0 / 1,0) + 0,919 ∙ 311,6 / (1,0 ∙ 755,76 / 1,0) = 0,418 ≤ 1,0 OK

97,6 / (0,633 ∙ 4183,0 / 1,0) = 0,037 ≤ 1,0 OK

Deformation

Horizontal displacement of top point of column:

0,002 m

Accepted value:

H / 150 = 0,036 m

D / Dacc = 0,056 < 1,0 OK

metal structures design project iii steel hall examples of

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